text:
How to read a TAF report which looks like the following?
How to read a TAF report which looks like this:
TAF: VOBL 010800Z 0109/0118 09005KT 6000 SCT010 BKN080 TEMPO 0109/0118 3000 RA/DZ SCT007 SCT012 FEW025CB OVC080

What's your specific question? How to interpret the codes, or is there a code you do not understand?

http://www.flightutilities.com/MRonline.aspx

You should make basic research on how to read a TAF report and then precise what is unclear in this given TAF

You can find how to decode TAF at aviationweather.gov site.
A TAF report usually contains the following sequence of elements:

Type of Report
ICAO Station Identifier
Date and Time of Origin
Valid Period Date and Time
Forecast Meteorological Conditions

TAF- Type of report
VOBL- ICAO station identifier
010800Z- Date and time of origin
0109/0118- Valid period date and time
09005KT- Wind
6000- Visibility in metres
SCT010 BKN080- Thousand scattered, ceiling eight thousand broken cumulonimbus clouds
TEMPO - Is used to denote temporary conditions which may arise, followed by the forecasted meteorological conditions.
0109/0118- between these times
3000- Visibility in metres
SCT007 SCT012- Same as above


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For what program was this Cessna 182 fitted with canards?
I want to see if anyone knows what the heck is going on here. I was doing a flight out in Palo Alto California recently, and during the run up this Cessna 182 pulled in next to us.   It appears to have a canard foreplanes attached near the firewall of the airplane, and it appears to have a movable control surface on it as well. No this is not a Photoshopped image, this is an actual airplane that was sitting right next to us. I’m just curious if anyone here knows of any program using this aircraft and the purpose of the canard foreplanes.  Thanks.

It's a King Katmai STOL mod for a Cessna 182. From https://www.avweb.com/ownership/king-katmai-mod-safe-stol/

The full King Katmai modification consists of the canard, 300-HP IO-550, choice of 82- or 86-inch, three-blade prop, speed mods/drag reduction fairings, wing extension, increased gross weight, heavy-duty landing gear, which includes Cleveland brakes, stainless steel leading edges on the gear legs, brake lines faired in behind the main gear legs and routed so they are unlikely to be snagged during rough field operations, heavy-duty Airglas nose strut and oversize tires.

It brings the stall speed down to 31 knots. 3100 lb gross weight takeoff distance of 290 feet from a hard surface runway.

Yeah come to think of it that’s what was written on the nose of the aircraft. I never realized there had STOL kits featuring canards.  Thanks, Terry!

Could it be the Kenai, since it doesn't seem to have the large main gear tires?

Whoa, I thought the Grob 115C had a low stall speed at 45 knots with flaps at 60°.

Terry got it right!  It’s a King Katmai STOL Kit.  You can practically squeeze that thing into a parking lot!

He had me right up to the part where he said it cost $350K(!).

Yeah they basically do a spinner to tail rebuild.  Not exactly a run of the mill STC kit!

For those who do not know -- the difference between a King Katmai and a Kenai is whether or not the plane includes the Wing-X STOL wing extensions, which add about 3' to the wingspan.  AFAIK, everything else is configurable.  I cannot tell from the picture whether or not the plane has the wing extensions, so the picture shows either a King Katmai with small wheels (a bit unusual) or else a Kenai.  Note that the video's performance figures are for a King Katmai with its usual tires (29" Bushwheels), not the smaller tires with wheel pants shown in the picture.
(The above assumes it had the 300HP IO550.  If not, there is one more choice -- a 260SE.  Think of that as a Kenai with a 260HP engine.)
As to price -- it can go up to the mid-$400K's -- depends on things like the avionics requested, how fancy a custom interior you have installed, and how fancy a paint job you have done.

This would be a better fit as a comment than an answer.

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Is the APU classified as a component of Airframe or Powerplant?
In the classification of systems and components on the aircraft - is an APU classified under the Airframe or the Powerplant?

I was a tech writer in another life.  Under ATA100/2100, the Air Transport Association system of categorization of systems for maintenance manuals, the APU is Chapter 49 (Airborne Auxiliary Power).
Power Plant chapters are 70 and up. (Some OEMs also use a equivalent numbering convention for engineering drawings to align them to ATA100/2100, where, say, if Flight Controls are Chapter 27, flight controls related drawings would have "27" at the start of the drawing number with the applicable sub-system numbering as appropriate - makes finding drawings in a database quick and easy).
Airframe systems are Chapters 20 thru 50.  So from that standpoint, its Chapter 49 designation makes the APU an airframe system, not a Powerplant (which only refers to propulsion engines per ATA), as far as the industry is concerned, and I think you are pretty safe to go on that.

In the U.S., auxiliary power units are categorized under Powerplant for the purposes of maintenance. The FAA Aviation Maintenance Technician Handbook - Powerplant, Volume 1 lists auxiliary power units in the category of turboshaft engines on page 1-38.

It is a "power plant", so it would be classified as such. An APU isn't much different than a small jet engine.

Or it's part of the electrical system. The classification can go several ways.

@PeterKämpf I'd argue that the engine part is a power-plant, the generator part is electrical, either way the tech that has to work on it will be a power plant person, not an airframe one.

From both the design and maintenance point of view it is a power plant (see the answer about US maintenance standards). APUs from big jets have occasionally even been used as engines in some light aircraft.

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Is it possible to display WAAS enabled GPS Flightlines on older mechanical flight directors such as S-tec?
We are in the research process of a panel upgrade in a M5 Maule.
The ac is used for transportation and aerial survey to support ground survey. We frequently fly tight, low alt grids. Lateral and vertical deviation limits are 50’.
Is it possible to display WAAS enabled GPS Flightlines on older mechanical flight directors such as S-tec?

Not to my knowledge.  No.

I think you may elaborate a bit more the context. Moreover, try to use less abbreviations or explicit them (e.g. ac, alt grid).

Thanks, my first draft was quite lengthy and It seems that I over edited.

We have an occasional need to fly multiple parallel flight lines with a high degree of precision. We can purchase computer software designed for this purpose but its quite bulky, hard to mount anywhere near the flight instruments, and very expensive.

Avionics shops are quite busy these days and are not interested in low budget jobs. So we are looking at options. An EFIS system would be great but far too expensive.


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Can UHMWP be used as composites for aircraft impact susceptible areas?
Can Ultra High Molecular Weight Polyethylene (UHMWP), be used as composites for aircrafts impact susceptible areas?
I did some research but could only see UHMWP being used as a hybrid with graphite/ carbon.
I have two questions is it possible to make composites purely of UHMWP and epoxy?
And, why use UHMWP with carbon, as it already has good characteristics alone?

From what I've read it doesn't bond reliably to epoxy resin, at least reliably enough for aerospace standards. Mixing with carbon may be partly to obtain some of the benefits while relying on the carbon to bond to the resin.

Have you compared it with kevlar/epoxy composites? You know, since these are the current go-to solution for impact-prone areas.

Like Kevlar and UHMWP hybrid fibre mat? I've compared ti yet, since I was considering the possibility of a purely UHMWP fibre composite.

No, I mean if you have any indication that it would be any better than kevlar/epoxy. If even back of the envelope calculations point to kevlar/epoxy being better, it would explain why there has been no serious research into UHMWP composites for this purpose.

Yeah might be something that might rule out UHWMP as a pure fibre, I found out that it has low operational temperatures so might be the reason it's not used alone on its own.

@AEheresupportsMonica The density is below 1, so you can use a lot more when replacing glass- or carbon fiber. This improves strength and buckling stiffness, and pound for pound a UHMWP-reinforced structure will absorb considerably more energy.

Short answer: Yes, when used properly. It does have its limits (low melting point, low adhesion with epoxy), but if you design around those limits it helps to raise energy absorption considerably.
Long answer: I've used Spectra for improving the crashworthiness of glider cockpits in a study. That was back in 1987-88 when that stuff was brand new.
I molded models of forward fuselages in glassfiber-epoxy as a baseline and then added more models using graphite, aramide and Spectra woven fibers. UHMWP is great in tension but poor in compression, so using it all by itself for the fuselage cannot be recommended. I used 100% Spectra layups only for the seat pan of the models but covered it on both sides with glassfiber or graphite fiber for improved compression and shear strength when using it for the fuselage shell.
Next, I set up a dynamic and a static test. The dynamic test used a swing to let the models crash into an angled, greased steel plate while the static test used a hydraulic testing machine which slowly lowered a steel plate on those poor, unsuspecting models until they were completely crushed. Using a video camera and strain gages, I could prove that the quasi-static test delivered equivalent results to the dynamic test (since the speed of sound in the material is much higher than the impact speed), so the much easier to measure static test could be used for the majority of the tests.

Left: Quasi-static test in progress, right: Model mounted on the swing. Own work.
While the glassfiber and carbon baselines produced a strong force spike at the beginning of the impact and then developed cracks which greatly reduced the stiffness and impact energy absorption, the Spectra samples would not crack in the same way: The UHMWP fibers prevented those cracks. Instead, the epoxy matrix crumbled over large areas and the cover layer delaminated, producing a nice, continuously high force over the duration of the impact. I could improve energy absorption by 200%.

Lineup of some models (after the test). On the far right is a graphite shell with an aramide seat pan.
Those strain gages you see in the pictures were used to calibrate a FEM model using ADINA, a nonlinear code which allows for large displacements. Running the FEM model on a Microvax took days but produced very convincing results after I had measured the material properties of the materials employed in the models. Never use manufacturer-supplied data but gather your own!
The results were published in the Journal of Aircraft.

And... this is why you were our first 100k rep member and will be the first 200k rep member. Would you identify the construction of the other 3 cockpit models in the last pic? The 4th (black) one you identified as graphite & aramide, but the other 3 aren't clear (at least to me).

@FreeMan: Now is more than 30 years later and I did not record what I photographed back then. So this is from memory and error-prone: The leftmost one is 100% glassfiber, the one next to it is most likely a glassfiber shell with an aramide seat pan (because that was the structure of the ASW-20 C I helped to complete earlier) while the third one is most likely a glass-Spectra-glass sandwich for the shell and a Spectra seat pan. I could also be a glass shell with a Spectra seat pan. Spectra is almost white, so I am pretty sure concerning the seat pan. I just picked some from more than 10 models.

A guy could ask...

According to Wikipedia, its thermal properties are mediocre:

The weak bonding between olefin molecules allows local thermal excitations to disrupt the crystalline order of a given chain piece-by-piece, giving it much poorer heat resistance than other high-strength fibers. Its melting point is around 130 to 136 °C (266 to 277 °F),[8] and, according to DSM, it is not advisable to use UHMWPE fibres at temperatures exceeding 80 to 100 °C (176 to 212 °F) for long periods of time. It becomes brittle at temperatures below −150 °C (−240 °F)

which severely limits the choice of matrix for a hypothetical UHMWPE/resin composite.
In particular, the curing temperature of the epoxy resins typically used in aerospace applications are around 200°C, which would either require the use of a cooler-curing epoxy (with lower mechanical properties) or a different solution altogether.
So, as a partial answer to your question: yes, it could be used, but there does not seem to be a reason for it, as the readily-available composites have better properties for the use cases defined thus far.


text:
why does the flow need to decelerate (by a thermal throat) in a scramjet?
I have a question about the thermal throat in a scramjet. From my textbook:

"The scramjet, on the other hand, needs an area increase as combustion occurs. For this reason, on the scramjet the mechanical throat is substituted by a thermal throat that decelerates the flow through
tailored heat release."
why does the flow need to decelerate in a scramjet?
if flow in a scramjet is supersonic, wouldn't heat release
accelerate the flow? (from general 1d form of equations du/u=1/(M^2-1)*(dA/A+dF/P-dQ/CpT))
Where is the thermal throat in this flow located?

I found an answer, for those who are curious. The thermal throat described in a scramjet is not an ideal case. Essentially, when the flow is supersonic, heat addition done in the combustion chamber can slow the flow down to subsonic speeds (based off the 1-d heat velocity, Rayleigh Flow). When this happens the flow is considered thermally choked, and is a bad case for this engine. To prevent this thermal choke from happening, an the area of the combustion chamber increases with the direction of the flow (area increase increases velocity for supersonic flow).

You should post that as an answer, rather than a comment, so that it can help future visitors to the site.  Welcome to Aviation.SE!

I found an answer, for those who are curious. The thermal throat described in a scramjet is not an ideal case. Essentially, when the flow is supersonic, heat addition done in the combustion chamber can slow the flow down to subsonic speeds (based off the 1-d heat velocity, Rayleigh Flow). When this happens the flow is considered thermally choked, and is a bad case for this engine. To prevent this thermal choke from happening, the area of the combustion chamber increases with the direction of the flow (area increase increases velocity for supersonic flow).

i got the impression that mach number decreases also because speed of sound increases.

That should also decrease mach number, I think both effects would be taking place then.

i would expect a pressure increase for that. i'm used to thinking of combustors as constant pressure, constant velocity, with the increase in volume from combustion countered by increasing cross sectional area down the combustor.

Make sure you give yourself a check mark for the answer, so others know this has a solution.

Abdullah, that was an interesting point. I asked my professor about this. The pressure in a traditional combustor stays constant because there is a decrease in density as temperature increases. Since mass flow rate must stay constant, there in turn must be an increase in velocity to account for the decrease in density. From what I understand, the combustor velocity increases to keep pressure constant, not the cross sectional area of the combustor.

Strange......why would the flow in a constant cross section duct speed up rearwards without creating a force (pressure) in the forward direction (Newton's third law)?


text:
What does it mean for a door to be armed?
What does "arming a door" mean?  Is the orange tape that's applied over a window related, perhaps as an indicator to people outside that the door is armed?

Ever since 911 they want to make sure doors can defend themselves. The bottom line here is: don't mess with an aircraft door, because they know how to handle themselves.

@Tyler Your username is really appropriate for that comment.

@SteveJessop: Nobody trusts the windows anyway after _NSAKEY was found.

@Honza - I appreciate the edit suggestion, but do you have the source/credit for the picture?

@JamesThorpe Sorry for the late answer - I took the picture myself with my phone.

If a door is put into armed mode, it will trigger the evacuation slide when the door is opened.

Prior to departure (usually before engine startup), all the aircraft doors are placed into the armed (or automatic) mode by the cabin crew. Methods of arming vary from aircraft to aircraft, but ultimately the girt bar (a metal bar attached to the door end of the slide) is physically attached to brackets either in or adjacent to the door sill.[...]If a rapid evacuation is required and the doors are opened while "armed", the opening of the door pulls the slide pack out of the bustle (because the girt bar is physically attached to the aircraft floor)

This is also the reason for the famous All doors in flight announcement:

DOORS TO ARRIVAL AND CROSSCHECK “Flight attendants, doors to arrival and crosscheck.”
Meaning: Occasionally heard as “disarm your doors and crosscheck,” and announced by the lead flight attendant or purser as a plane approaches the gate. The intent is to verify disarming of the emergency escape slides attached to the doors. When armed, a slide will automatically deploy the instant its door is opened. Disarmed, it needs to be deployed manually. On departure the slides are armed to facilitate an emergency evacuation. (You might hear this as “doors to automatic.”) Upon docking, they’re disarmed to keep them from billowing into the boarding tunnel or onto the apron during servicing.

The orange or yellow tape is a visual indicator for ground workers or ramp agents that the doors are armed or not armed. The visual indicator can vary, it can also be built into the door lock mechanism, triggering a visual indicator externally if the door is locked and armed. In the following picture, the orange tape is not strapped over the window, which should indicate an unarmed door. In the video that this image is taken from, the flight attendant has just locked and possibly armed the door, the other flight attendant would place the orange tape over the door window during "cross-check".

(Image Source: YouTube "B757-200 Door Closing" - Author: Exupery747)

Well out of what I really know about — but from memory, around Europe, it's always one of the pilots who announces changes to the door's armedness.

Airline Ops dependant, can be flight deck or purser.

When the door is armed, the slide will deploy when the door is opened.
The tape is indeed an indicator to people outside that the door is armed.

Can somebody from the 'outside' open an armed door?

@Firee Sounds like that should be a new question, with the potential for some "amusing" videos...

@Firee For example in this A321 video (from 2:40) it shows that opening from outside automatically disarms the door. https://youtu.be/Vbm8MpC0e1o

Yes. Someone can (and unfortunately has) opened an aircraft door while still armed. Some aircraft have an indicator that is visible from the outside indicating the door is still armed. Other aircraft require indincation from flight attendants (such as a thumbs up) indicating the door is disarmed. Older single isle aircraft  (727 DC-9) the slides would fall out of slide pack but not inflate. Flight attendants would have to pull an inflation handle at the part of the slide attached to the floor of the aircraft after opening the aircraft door. This was an economic feaure for airlines as slide deployment are expensive. If the slide deploys but not inflated they were just rolled back into the protective covering.  However these were dangerous in actual emergencies as flight attendants would have to reach down and inflate slides and could be pushed out. The ones that automatically inflate are dangerous if opened accidentally if they are deployed into a jetwat or catering vehicle

IIRC the Airbus training video shows that opening the door from outside disarms them as part of the process. I would expect the newer Boeing types to work the same.

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Are the A400M propellers rotating in the wrong way in "Mission Impossible: Rogue Nation"?
At the beginning of the movie Mission Impossible: Rogue Nation we can see an A400M taking off.

When the engines start, the propellers turn in opposite directions. As I understand it this is not a common configuration but correct for the A400M according to Wikipedia. However, the Wikipedia article also says that the configuration is "down between engines (DBE)" which is opposite from what we see in the movie.

Did the movie get it wrong or did they use a special/early version of the A400M which worked differently?
I'm also a little confused about the propeller blades, do they have variable pitch?

Welcome! Thanks for jumping in with an interesting question. I’m not at all familiar with the A400M, but I’m sure one of these other geniuses will get you sorted out quickly ✌. Hope to hear more from you soon.

Tomas I edited my answer based on your comment, it may very well be that the scene simply is played bacwards, I explain the propeller "logic" in my edit. Good thinkin on the reverse scene, didn't think of that myself

This is two differnt question and they should be asked separately. BTW, a quick basic research shows they have variable pitch.

I have seen the movie, but it failed to occupy my memory... anyway:
Yes, the A400M propellers are contra rotating, and more specifically DBE as you described. I am not 100% sure, but I'm willing to bet all A400M's ever produced have this configuration.
So if the propellers seem to rotate in the "wrong direction" there are two possible causes:

it is an effect caused by camera shutter speed in relation to propeller rpm. This may make the propellers seem to rotate in ever which way, or even stay still, as you can see in this youtube video.
the whole scene is cgi, so there actually is no real plane involved. My money is on this...

The propeller blade angle in A400M, as in any larger modern propeller aircraft is adjustable.
This type of propellers, or  the governor adjusting the blade angle (pitch) of the propeller to be precise, are typically constructed in such a way, that when the engines are off, there is no oil pressure, and the blades are set in feathered position. So when the engine starts up they will be feathered, and when the engines are shut down, the propellers will adjust to feather. This is basically a safety feature to minimize drag in engine failure situation. To prevent thrust from forming during and immediately after start, the propeller adjustment is set to feather by pilot, this prevents the propeller angle change as oil pressure in the system rises and governor would otherwise start to adjust the blade. As you commented below, it is totally possible the scene is played in reverse, as the blade position would be the same in both shut down and start.
The picture you posted in the question shows propellers in feathered position, so they are not producing thrust. If this is a screen capture from a scene where the plane is taking off, it most definitely is cgi. With propellers feathered, the plane will not be gaining speed, and definitely not taking off.

Thanks. I have discounted the camera shutter speed option because the propellers can be seen spinning up slowly from 0 rpm.

Wait, are they really spinning up? Maybe they are spinning down but it's just played backwards. The feathered position may have been used to keep the camera & crew safe in front of the wing. In any case, Tom Cruise was actually on that plane, a lot of it was real.

Surprisingly, they did strap Tom Cruise to the outside of an A400 taking off while the crew were filming on the ground: https://www.youtube.com/watch?v=p3XgoDa7RB4

Nice find! In that video, at 1:37 (https://youtu.be/p3XgoDa7RB4?t=97) you can see the propellers spinning, and although they appear to go UBE, you can see the blades are actually bent to go the other way, wich means that's probably a strobe effect. That's consistent with the way they're twisted in the scenes where the aircraft is still. So the actual aircraft used for the filming is DBE. -- But the plastic model is different, and sketch at 0:23 shows all propellers going the same way. -- so it's probably a CGI issue

As a sidenote, mr Cruise hanging outside an actual plane is a (clever, I admin) publicity stunt. Current CGI is so advanced, you just can't tell the difference. It would have been cheaper, safer and faster to fake that, but when you count in the publicity value both for the movie AND mr Cruise, it makes sense. Sort of


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What does induced flow exactly means on helicopters flight? How is it different from downward deflection of air?
I believe all airfoils deflect air downwards at an angle at a velocity V. What is the difference between the downward-vertical component of the deflected wind and the induced velocity in Helicopter flight?
Corollary: Why the airplane doesn't face this induced velocity and induced drag?

These are referring to the same thing.  The induced velocity is the change in air velocity due to aerodynamic interaction with an object.  The downwash from a helicopter is the vertical component of the main rotor's induced velocity.  This is the induced velocity of a bunch of rotating wings.
A wing on a fixed wing aircraft produces induced velocity too, but it's located in a region of lower induced velocity during flight.  The induced velocity varies in space and time.  As a fixed wing induces velocity in the flow field, it "flies away" from that region.  So it's flying in "cleaner" air.  An exception would be if a fixed wing could fly in a tight circle.  In this case it would fly through more of its induced velocity, experience more induced drag like a helicopter, and have a large downwash below the circle.
Another way to say this is that the fixed wing's induced velocity is more spread out in space, and hence not as concentrated at any one point.  The hovering helicopter's induced velocity is highly concentrated in the region around the helicopter.
You can probably find visualizations or animations of vortex lines on the web.  In those, you will see that a fixed wing primarily trails vortex lines behind it, while a hovering helicopter produces a helix of vortex lines below it, but close by, which keep the induced velocity at the rotor disk high.

Perfectly explained. Thank you.


text:
How hot is supersonic intake air?
How hot is the air at the compressor inlet at supersonic speed? For instance the Concorde at mach 2.2 or other jet engines at even higher speeds.

Are you interested in how the temperature varies with speed for a specific aircraft or in what is the maximal intake temperature for several supersonic aircraft designed to cruise a different speed (Concorde, SR-71, F-104, Mig-31,...)?

The air entering the first compressor stage of the SR-71's engines could be as hot as 450C while at maximum cruise conditions. If that air got any hotter, it would begin to soften the compressor blades, and this put an upper limit on the speed of that plane.


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What can be approach of balancing canard glider model?
There are a lot of instructions out there about balancing classic wing configuration glider models. The center of gravity should be approximately 25% to 30% from the front edge of the wing, which is easy to control. The next step is flight tests and adjustment by adding or removing weight or moving the wing forward or backward. Using this approach I constructed several gliders with stable horizontal flight.
But what can be approach for balancing canard configuration glider model? Is it the only recommendation for the canard wing to produce more pitch applied to the center of gravity then the main wing produces, is it possible to verify this somehow before testing flights?

Static longitudinal stability is defined as the tendency of the airplane to pitch down when it slows down and vice versa. In other words, when the lift coefficient goes up, the pitch moment coefficient must go down. This is expressed in equations like $$-\frac{c_{M\alpha}}{c_{L\alpha}} = \frac{x_N - x_{CG}}{l_\mu} = X\%$$ with X a positive number between 0.1 and 0.2. $x_N$ is the lengthwise coordinate of the neutral point, counted positive backwards, and $x_{CG}$ is the lengthwise coordinate of the center of gravity. The neutral point of both surfaces is at their respective quarter point and the combined neutral point sits between both, proportionally closer to the larger surface. Place the center of gravity X% of the mean aerodynamic chord $l_\mu$ ahead of this common neutral point and your stability is X. 0.2 is a rather high value and 0.1 makes for better performance but also less inherent stability.
Now all that is missing is that elusive neutral point of a canard. To explain what follows, it is the combination of the neutral point of both surfaces scaled by their area and corrected for the downwash of the forward wing which reduces angle of attack changes on the rear wing. Since only the inner wing will be hit by this downwash, we need to get a grip on the fraction of the wing which is exposed. This involves the span ratio of both wings $\frac{b_c}{b}$ and the taper ratio $\lambda = \frac{c_{outer}}{c_{inner}}$ of the main wing.
$$x_{N_{canard}} = \frac{x_{N_{canard}}\cdot c_{l\alpha_{c}}\cdot S_{c}+x_{N_{wing+fuselage}}\cdot c_{l\alpha_{w+f}}\cdot S_{w+f}\cdot\left(1-\frac{\delta\alpha}{\delta\alpha_W}\right)\cdot\frac{2+(\lambda-1)\cdot\frac{b_c}{b}}{(1+\lambda)}}{c_{l\alpha_{c}}\cdot S_{c}+c_{l\alpha_{w+f}}\cdot S_{w+f}\cdot\left(1-\frac{\delta\alpha}{\delta\alpha_W}\right)\cdot\frac{2+(\lambda-1)\cdot\frac{b_c}{b}}{(1+\lambda)}}$$
In absence of a precise value, use 0.7 for the downwash factor $\left(1-\frac{\delta\alpha}{\delta\alpha_W}\right)$.

Exactly the same as a conventional wing/tail.  You use the combination of lifting torques around the center of gravity to balance at the center of gravity.  With some planes this is as easy as detaching the horizontal stabilizer from the back and putting on the front, with up lift instead of down force.  Very appealing to the eye but ...
You now must evaluate stability, especially in pitch.  Essentially, a canard reverses the role of tail and wing, you now have a much smaller fore wing, and the main wing (tail) must provide stability.  With deltas, this is very practical, giving their large trailing edge area.  With a straight wing, the results can be disasterous, as seen on the USAAF Ascender.
A lower aspect wing would be more stable with a canard, but the conventional high aspect wing sailplane design with a rear mounted horizontal stabilizer is hard to beat.

Was "straight wing" the reason of USAAF Ascender crash or crashes, is there any additional information about this?

There design actually was slightly swept, and did ok until they moved the weight back too far.  More area was added aft of the CG.  This type of design had better luck when fitting into a supersonic nose shock wave was required.
Subsonicly, the "classic" design works better for a number of reasons.


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How are coefficients of equation of sideslip angle in the paper "A general Solution to the Aircraft Trim Problem" calculated?
I am sure many of you guys must have read the paper, "A General solution to the Aircraft Trim Problem" by Marco, Duke and Bernt. I am working with the turning of the Aircraft and I am not able to understand the calculation of Beta (sideslip angle) given in equation 72.
What I am able to understand is that the author describes the wind frame and body frame by 2 simple plane equations and then takes a dot product of the two planes to calculate beta.
The problem I am facing is with the calculation of the coefficients of the plane. They say that the coefficients can be expressed according to equation 67, which is nothing but axis transformation equations. Can anyone help me with the detailed calculation of those coefficients.
I have attached the above mentioned part as link to image file. Please have a look into it.

Link: A general solution to the Aircraft trim problem

If you don't have an answer here in few days, you can try on Physics.SE. They are more used to read research paper and to deal with forces and frame of reference changes in Newtonian mechanics.

First, some note on this paper: from my cursory reading, it always assumes zero atmsohperic wind relative to the ground. That's why the wind axis represents the true motion of the aircraft. In general, this is not true, therefore $\psi_W$ and $\theta_W$ are not the track and flight path angles.
Now onto your question. When you consider zero wind, then the angular difference between the y wind axis ($\hat{j}_{W}$) and the y body axis ($\hat{j}_{B}$) is the sideslip angle, $\beta$. This is consistent when you consider flow incidences as 3-2-1 Euler rotation from the body axis, where yaw (sideslip) is the last rotation:
$$C_{BW} = \begin{bmatrix}\cos\alpha\cos\beta & -\cos\alpha\sin\beta & -\sin\alpha \\
\sin\beta & \cos\beta & 0 \\
\sin\alpha\cos\beta & -\sin\alpha\sin\beta & \cos\alpha
\end{bmatrix}$$
Mathematically, this can be immediately seen when you rewrite the rotation matrix in its equivalent direction cosine form:
$$C_{BW} = \begin{bmatrix}\hat{i}_W \cdot \hat{i}_B & \hat{j}_W \cdot \hat{i}_B & \hat{k}_W \cdot \hat{i}_B \\
\hat{i}_W \cdot \hat{j}_B & \hat{j}_W \cdot \hat{j}_B & \hat{k}_W \cdot \hat{j}_B \\
\hat{i}_W \cdot \hat{k}_B & \hat{j}_W \cdot \hat{k}_B & \hat{k}_W \cdot \hat{k}_B\end{bmatrix}$$
So: $\cos{\beta}=\hat{j}_W \cdot \hat{j}_B$.
The gradient of the Cartesian equation of a plane is its unit normal, when normalized. For the $x_W z_W$ and $x_B z_B$ planes, the normal is simply their y axis (i.e. $\hat{j}_W$ and $\hat{j}_B$, respectively.
Lastly, eqn (72) can be worked out when $\begin{bmatrix}x & y & z\end{bmatrix}^T$ in the planar equations are in the inertial frame ($_V$), and transformed to the respective frames. The coefficients can then be derived and massaged to obtain (72).

The angular difference between wind axis and body axis is the side-slip angle even if you consider non-zero wind. There is not much point in considering non-zero wind unless the aircraft is taking off or landing anyway—the ground is not relevant, so you don't ever need to calculate in the ground reference fame at all.

@JanHudec Yeah, I've never said the winds make any difference to sideslip. They, however, make all the difference to the track and flight path angle, which receives a lot of attention in the paper.

Hey buddy thanks for the answer, but I already understood the part that you tried to explain , I am facing more issues in deriving the coefficients of eqn 72. Can you help me with the maths regarding that? Or just guide me through the steps? Thanks


text:
Would fuel quantity indications be affected by extended flight at nonzero bank angles?
If flying with a 20kt crosswind requiring yaw input as well as roll input to maintain course, would the fuel indicators show different loads depending on the roll angle of the aircraft (in this case, an MH-53e)
Clarification: With a left bank angle of 15-20 degrees would it appear that the right side fuel cells were using fuel faster with a quantitative fuel quantity system?
Thank you sir. That is a better way of saying it.

Isn't the plane still flying straight and level even with a crosswind, even though course and heading are different?

It depends, what indication are you looking at? I'm guessing on the MH-53 it is a fuel totalizer, not an actual gauge. If it were a physical gauge it may change, but more expensive military/commercial aircraft use totalizers.

Crosswind requires a 20deg bank angle with yaw input to fly along a predetermined course. It's pretty windy and this is an actual pilot reported issue.

4 tanks and it is a totalizer system so each probe covers the tanks from top to bottom and you can transfer to and from all tanks

Where did you get the idea that flying a predetermined course in a crosswind requires a constant nonzero bank angle?

While towing mine countermeasure gear pilots must fly a predetermined track and pilot reported having to fly at the angle

@Steve compensation for the wind needs yaw normally

@Derek You should learn the difference between crab (putting your nose into the crosswind) and side slip (lowering your wing into the crosswind). The second thing is probably never used in steady flight, it's usually used only in final moments of a crosswind landing and in first moments of a crosswind take-off.

I do actually know the difference.  The aircraft was towing minesweeping gear and required both yaw and roll inputs to follow the track that was being cleared.  The tracks are plotted by gps coordinates. This is an actual pilot reported issue which I was the primary troubleshooter. We call it crab walking and unless you understand that they are dragging an object through the water while flying under 100 ft which not a normal thing most pilots have done you shouldn't be condescending in your reply yo'

I don't know what fuel quantity indicating system is installed on the MH-53, but most aircraft use a capacitive indicating system. Basically, there are multiple probes spaced throughout the fuel tank(s). During a turn or bank, the fuel will accumulate to one side of the tank, which increases the capacitance on the respective probes on that side of the tank. Meanwhile, the capacitance decreases on the probes that aren't covered as much. Total capacitance of all the probes is then summed and processed. Basically, no matter how the aircraft is oriented, in a properly operating system, there should be no detectable change in fuel quantity.

It is capacitance based but has a single probe per cell and bags out around 24k pounds split between 4 cells

Each cell has its own indicator plus a total just to clarify

Generally speaking, a capacitance based system will be designed so that all but the most extreme flight characteristics are accounted for. My guess is that there would be very little or zero change. That said, it's possible there is something in the flight manual or maintenance manual about extreme flight characteristics and the fuel qty reading.

As the overall load decreases the appearance of a miscompare increases. Even though fuel is being used from each tank equally the stationary fuel probes show more fuel on the wing down side.

Are you sure fuel is being used from each tank equally as the level decreases and these maneuvers are performed?

@Frank I'd add that in violent maneuvers / turbulences, you simply wouldn't rely on the fuel readings for a while if they started behaving weirdly.

@yo' I'm thinking of this from a strict maintenance standpoint, as I'm not a pilot. I am presuming these maneuvers to be for an extended period of time rather than a quick maneuver. I would agree that in extreme maneuvering, it is possible for there to be some unreliability. That said, if a maneuver can be held for a long period of time, it's not so abnormal is it?

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
How does the forward speed of an auto-gyro relate to the rotation speed of the rotor during auto-rotation?
In my other question I was looking for a relation between the forward speed and the thrust (lift) of an auto-gyro.
I have come to the conclusion that for an auto-gyro in auto-rotation the forward speed will rotate the rotor and with that regulate its rpm. The faster an auto-gyro flies the faster the rotor rotates resulting in a higher thrust.
What I hope to accomplish is forward braking where the auto-gyro flies faster than needed to provide sufficient thrust, resulting in a higher rotation speed which will then result in more thrust and lift the auto-gyro up. This additional thrust isn't needed as the same level is required. By braking the rotor, using a generator, the rotor speed is reduced till the point it is still able to maintain level using auto-rotation. For this theory I need to understand the relation between the forward speed and the rotation speed.
But it is still not completely clear how this forward speed relates to the rotation speed. I hope someone can explain it to me or point me into the right direction.
I might be completely wrong and came to the wrong or inconclusive conclusions if so don't hesitate to tell me.
Thanks in advance!

An autogyro flies at constant rotor RPMs. In vertical autorotation, that is a particular case, the rotor revs may be 10-15% less, but in straight and level flight, the rotor revs are higher, and always the same, independently of the speed of the gyro. It cant be otherwise, since, if the rotor revs went higher with the speed, the rotor thrust would also rise, and the gyro will gain altitude, no longer flying straight and level...
It's also true that if, flying s/l, the pilot opens the throttle, he also pushes the stick slightly forward in order to reduce the AoA of the disk, and thus keep flying s/l...


text:
What if you approach the VOR station such that your turn OB is not on the holding side?
I'm reviewing the Sporty's ground school for VOR approaches, and their example shows the approach to the VOR station then heading OB as tracking on the holding side. What if you approach the VOR station such that your turn OB is not on the holding side? I know conventionally that side is not the protected side of a hold, but does the hold side matter for that timed turn? Are you typically just vectored such that you approach the station and turn OB on the protected side of a holding pattern?

I'm so stale on this stuff, but let me take a stab. My first impulse is that you're assuming that the pilot would execute the same type of turn to establish OB regardless of the flight path's resulting orientation to the hold. Ensuring the aircraft remains on the protected side is always goal. So instead of a direct entry to the OB as depicted, wouldn't you want to execute  either a teardrop or parallel entry to the hold, which usually terminates about halfway through a normal hold entry once the aircraft is oriented to allow you to establish OB? Does that make sense?

The picture and your description do not really match. The hold represented is a standard, right-turn holding pattern. It is probably meant as a holding pattern of the missed approach. It is not meant as a holding pattern for before you travel outbound for your procedure turn. Based on the standardized format of IAPs, the right-most triangle is the Final Approach Fix. The left-most triangle is the Intermediate Fix. The VOR is an Initial Approach Fix. If, for some reason, you have to hold above the VOR prior to joining the approach, do so at or above the IAF altitude and intercept the outbound radial by continuing your turn after the outbound leg. I would warn against holding over VORs without ATC providing aircraft separation. VORs are busy airspace.
The IAF will have an altitude associated with it at which you can approach the IAF from any direction and be clear of obstacles. If one is not present, approach the IAF at or above the MSA. If that were not the case, the IAP would have a notation designating the IAP Not Applicable/Available coming from certain directions. For instance, flying from left to right on the picture, ATC would clear you to start the IAP at the IF (which can be designated on the chart as an IF/IAF) instead of the IAF (VOR). Aircraft flying from all other directions would fly to the VOR to start the IAP. The IAF, procedure turn, IF (IF/IAF), and FAF would all have their respective altitudes assigned to them in that order. Your profile view would start with the altitude at which you would need to stay above in order to start your procedure turn. If not, start at your MSA and descend down to the PT altitude.
Something else that comes to mind is checking the sectional and charts for the MSA and whether the IAF is a mandatory fly over fix. If it is not, turning onto the indicated radial in a way to intercept it prior to crossing over the VOR would expedite your IAP. Depending on your speed and original heading, you would start your turn a quarter to a full mile out before reaching the VOR or radial.
Also, “the protected side of a hold“ is a misnomer. There is no such thing. There is a holding-side and a nonholding-side. The entire area around the hold is protected airspace based on the dimensions of the hold. See the following question:
what are the dimensions of the protected space provided by a holding pattern

I need to read some books. I'm uber-stale on all that, but just renewed that CFII!!!!


text:
How will LiFi application in an aircraft change the security of an air travel?
In a near future, LiFi technology will replace WiFi in a commercial aircraft. LiFi will make nonstop internet service possible. This will generate some new issues in air travel security. Especially security issues, let's say a terrorist group just needs to bring their notebooks or any other small computers with them in the cabin. They will be able to hijack the plane freely while they sitting down in the cabin. They can change the direction of the plane to whatever they wanted to, shut down the engine at any time, change the fuel amount, or even they made a ransom scenario. So, my question is:

Did the FAA already create a regulation to prevents such scenarios in the future?
Should the passengers be allowed or unallowed to bring their own computers? if unallowed, what will be the benefits of LiFi compared to the WiFi?

I fail to follow your reasoning: aircraft devices are not directly connected to internet and passengers systems (such as entertainment systems) are physically separated from aircraft instruments. Having a new way to connect passengers to the outside world (such as when wifi-based internet access was introduced in airliners) does not change anything to those premises.

@Manu, actually segregation between the 3 network domains (aircraft control, airline services, passenger services) is not physical, but logical. All domains need to communicate thru the same satellite link, so there are gateways. The risk has been ignored until recently (I've seen more technical descriptions of vulnerabilities, but can't put an hand on them at the moment). I'm not saying WiFi replacement is a factor in the whole picture.

@mins I don't understand why aircraft control need to communicate using a satellite link.

@Manu: This page give an idea (excepts the success of steering the aircraft from the IFE by Chris Roberts is a myth in my opinion). If you look at the top of page 6 of this study by Berkeley researchers, you'll have an idea of how domains are segregated logically.

@mins I'm quite surprised. I would expect to find at least one mention of vlan (which is used in my domain). I'll read this document and I may come with more precise questions.

I don't understand why you think a different connectivity technology would suddenly open up the aircraft's systems. Planes already have wifi onboard and, unsurprisingly, you can not control the aircraft using it.
I'm also not sure why you think Li Fi would change the quality of service. It's currently somewhat unproven but appears to be suited for short range applications, not long range uplinks.
In short, I think both the premise and conclusion is flawed and I can't see why security measures would need to change for LiFi

So, whatever the internet access method there will be no issue in security because FMS has a different communication protocol?

@Willysatrionugroho, it's not about protocol, but about physically separate communication medium and tight control over information flow between the separate networks.

+1  It doesn't matter whether I drive or walk up to your house, it's still the door that stops me getting in

Oh, I see it's not connected to anything it's a closed system. So, we should afraid of someone that able to make the key to getting into our house.

@Willysatrionugroho: Avionics is not a closed system, for example some instrument data are sent to a reporting system (ACARS to name it) which uses a satellite link to contact the airline or the engine manufacturer while flying. The entertainment system also uses this satellite link to connect the WiFi access point to an IP point of presence on the ground. Those connections are bidirectional and packets destinations, if spoofed, can be exchanged so the two domains now communicate. Of course security will try and hopefully succeed to prevent spoofing.


text:
Where do the compressed gasses of a charged engine goes?
Good day to you here.
I'm working on a modified version of 2-stroke engine. My plan is to use a supercharger for the scavenging process, instead of the conventional Crank case scavenging technique, thereby excluding the Crank case from the entire scavenging process.
But I have a question that's bothering me, In a charged (super or turbo) four stroke engine, after the the inlet port has been closed by the inlet valve, where does the subsequently compressed gasses go? Do they find their way back through the air filter? Do they remain in the inlet manifold until the inlet port opens again? Does this part of the engine cycle has any negative effect on the overall performance of the engine?
Note: Do excuse my grammatical blunders.

As long as the rpm is kept up, the charge coming out of the compressor has quite a bit of inertial mass and is more or less flowing continuously even though there is a valve downstream opening and closing.  There will be a pressure and velocity fluctuation near the valve as the charge just upstream is admitted to the cylinder, then cut off.  But because the opening and closing is happening so fast, these fluctuations taper off up stream.
This is one reason for having long intake runners on a 4 stroke; to take advantage of the inertial mass of the long cylinder of air in the intake duct.  On a non-supercharged engine, this helps with the preliminary filling the cylinder during the intake stroke because the piston doesn't actually have to "suck" initially.  As the intake valve opens, air comes rushing in pushed along by the inertial mass of all the air back up the intake duct, like a crowd of school kids jamming up in the hallway and piling out the door when it's opened at the end of school.  The ones closest to the door don't have to be pulled out; they are pushed out by their mates behind.
Anyway, the main issue is managing the intake charge and its inertial effects, and the increasing difficulty in doing this at lower speeds or idling.  I would expect you would have to do a lot of experimentation with compressor gearing/speeds and intake runner length trying to get an engine to run smoothly across the full operating range.
Your project sounds like a variation of a 2 stroke diesel, which doesn't use crankcase pressure to drive air into the cylinder like a spark ignition 2 stoke, and therefore requires a supercharger or turbocharger compressor to push the charge along in the absence of a 4 stroke's intake stroke to pull air in.  Possibly looking into 2 stroke diesels would help.
I'd also look into using an electrically driven supercharger in your design which would make it easy to experiment with different speeds and could really improve operation over the speed range by using a computer to manage it.

Making the charger electrically driven actually is part of my plan. I guess a 12v DC motor driving an impeller would be a good fit.

A two-stroke, opposed-piston diesel with scavenging blower was the Jumo 205, used as aero engine in the 30s and 40s... https://en.wikipedia.org/wiki/Junkers_Jumo_205


text:
What are some first-order drag approximations for airframe/propwash interactions?
What are some first-order drag approximations for airframe/propwash interactions, both for dynamic flight and for static thrust, in a traditional tractor configuration?

There is surprisingly little literature on that. This answer by @xxavier lead me to a very interesting paper in which the drag of a Luscombe 8b with and without thrust was compared. A very thoughtful way was used to arrive at zero-thrust numbers, and the result showed that the additional drag from the propeller effects was about 30% of the total zero-lift drag without thrust.
In the paper, the prop efficiency is 81% while the overall efficiency drops to 62% due to slipstream losses.


text:
What is the Function of Pressurization Light on Exit Doors for Commercial Airliners?
I am well aware that the average human is unable to open plug doors in flight due to pressurization of the cabin. I've noticed, however, a "cabin pressure" warning light in the window of exit doors, and specific instructions to not try to open the door when the light is on. Is there an actual danger of someone outside the aircraft being able to release something in or around the door while the cabin is under pressure, or is the light simply to indicate why they would be unable to move the handle?

Many doors are not plug-type these days though. Airbus doors open outward.

Doors opening outwards does not mean they’re non-plug. Both Boeing and Airbus use outward-opening plug doors. They only exception I can remember is the Boeing 767 which has inwards-opening plug doors for the main doors.

It is there to prevent "violent door openings":

(figure 1b source)

In some abnormal cases, outflow valves can remain closed, when the
  aircraft is on the ground, causing the air pressure in the cabin to be
  higher than the ambient air pressure outside the aircraft (fig.1b). In
  this case, there is a risk that an aircraft door could violently open
  and injure the operator or damage the aircraft.

(fig 2 source)

The Residual Pressure Warning Light (fig.2), part of the residual pressure
  warning function, warns anyone who wants to open a door if the
  aircraft is pressurized. This device is installed on all Airbus
  aircraft except for some A300, A310 and A300-600 where it was offered
  as an option. On these aircraft not fitted with the warning light, a
  caution placard located on the door reminds the operator of the risk
  that residual pressure may cause violent door opening.

This brief from Airbus has a lot of good info on residual cabin pressure as well.

Not to be confused with the "Slide Armed" light.

Huh! Unlike the slide warning indicators, I always assumed the light was just for emergency access, and not something used at the gate. Makes sense! Especially helpful that there's a logic diagram for when it's active on that linked page, because I noticed that it never illuminated on normal flights so I assumed it only activated when doors were unlocked or something.

People don't realize that a "small" pressure difference can result in a very big force, over an large area like a door. The warning pressure level of 2.5mb gives a force of abut 5 pounds per square foot. The total force on a pressurized door could be 100 pounds or more, which is not something the door-opener wants to discover unexpectedly when the latch is released.

In 2000 there was an incident where a flight attendant opened a door before the plane fully depressurized and was sucked out. He fell to the ground and died from his injuries.

Is there a real danger? Yes, there is. A private jet pilot was killed when hit by a door at Kittilä, Finland, 2018.

The aircraft had arrived in Kittilä two days before and was about to
  take off for a positioning flight without passengers. During these two
  days the aircraft was parked outside. Considering the season, the
  weather was usual. On the day of the arrival it was lightly snowing
  and the temperature was -5 °C; on the day of the accident the
  temperature was -22 °C.
On the day of the arrival the captain completed the final cockpit
  procedures alone while the co-pilot was outside putting on the
  aircraft’s external engine and sensor covers. At this point, the
  captain, apparently, closed the outflow valve because of the blowing
  snow.
On the day of the departure the co-pilot placed the aircrew’s baggage
  into the rear baggage compartment and began to remove snow. The
  captain and the cabin attendant boarded the aircraft. The captain
  started the APU which generates bleed air for heating the cabin and
  electricity for aircraft systems. The captain selected APU bleed air
  to be ducted into the cabin. Following this, the captain went outside
  to assist the co-pilot in removing snow and frost.
A moment later the captain went back inside to fetch a pair of gloves.
  When he came back out, he closed the door. A little later the cabin
  assistant inside the cabin felt pressure in her chest and ears. She
  went into the cockpit and knocked on the window to get the attention
  of the pilots. The pilots discontinued the snow removal. The captain
  opened the door which, owing to the significant differential pressure
  between the cabin and the outside, blew open with excessive force,
  hitting the captain and knocking him to the ground. The pressure wave
  also knocked the co-pilot down.

Full investigation report:  https://www.turvallisuustutkinta.fi/en/index/tutkintaselostukset/ilmailuonnettomuuksientutkinta/tutkintaselostuksetvuosittain/2018/l2018-01ilma-aluksenpaallikonkuolemaanjohtanutonnettomuuskittilanlentoasemalla4.1.2018.html

"Is there a danger" is in relation to the ability of a human to open a plug door while under pressure. Accepted answer brings up residual pressure, which is the case where the plug door could be opened while "pressurized". While this is an example of violent door opening due to pressure while on the ground, it is unclear that this example addresses the question of plug doors.


text:
What are various specifications of Engine oil used in Piston and Turbine engine aircraft?
I only know of two types MIL-O-7808 and -23699. Is there anything else ?

Some evidence of research could improve this question, but it's similar to existing questions like: What are the differences between fuel types (comparing with vehicles)?

@pritam the oil specifications are on the Jet Aircraft Engine Lubrication System link in my answer below. Just click on it and scroll down to the list of lubricant types there are about a dozen.

Here is a general answer for a general question.
Function of oil is three-fold for both piston and turbine engines:

Lubrication - reduction of friction;
Cooling - carrying away of heat; and
Cleaning - carrying away of contamination.

Oils used in jet engines must be able to withstand temperatures that are much hotter than piston engines.
Here is a link to some of the lubricating oils used in navy jet aircraft and the various specifications.
Jet Aircraft Engine Lubrication Systems
Piston aircraft engines tend to be air-cooled and are therefore manufactured with larger tolerances. This results in an engine that uses up oil (burns it). Most POH of piston aircraft that I've seen call for ashless dispersant oil, which refers to an oil that has non-metallic additives and has the ability to suspend combustion particles.
Here is an article that describes piston aviation oils.


text:
How long does it take for pilots to become unconscious after an uncontrolled depressurization at FL 340?
I've just read a news article on the depressurization incident that happened today in Australia. How much time did the pilots have before they lose consciousness if they hadn't used their oxygen masks? This incident is said to have happened at FL 340.

You may find some good info in this video about Hypoxia :
https://youtu.be/kUfF2MTnqAw

Similar Would failure to put on an oxygen mask during loss of cabin pressure result in death? and How long will one remain conscious in the event of a sudden cabin depressurization?

Just to point out, this doesn’t appear to have been an explosive decompression. So the actual cabin altitude probably didn’t exceed maybe 15,000 ft because the pilots made a quick descent.

According to: Time of useful consciousness:

Altitude (measured barometrically) - TUC
FL150 (15,000 ft; 4,550 m) - 30 minutes or more
FL180 (18,000 ft; 5,500 m) - 20 to 30 minutes
FL220 (22,000 ft; 6,700 m) - 5-10 minutes
FL250 (25,000 ft; 7,600 m) - 3 to 6 minutes
FL280 (28,000 ft; 8,550 m) - 2.5 to 3 minutes
FL300 (30,000 ft; 9,150 m) - 1 to 3 minutes
FL350 (35,000 ft; 10,650 m) - 30 secs to 1 minute
FL400 (40,000 ft; 12,200 m) - 15 to 20 seconds
FL430 (43,000 ft; 13,100 m) - 9 to 15 seconds
FL500 (50,000 ft; 15,250 m) - 6 to 9 seconds

Comments are not for extended discussion; this conversation has been moved to chat.

Time of useful consciousness tables are useful as a first order approximation.
The reality is that there is a great degree of difference between individuals, and also their conditioning. In one chamber ride, nearly an hour after explosive decompression (which can be very disorienting, even when anticipating it), I was still performing the cognitive exercises while my chamber partner was copying my kneeboard results and acting real silly.
I can also say that when spun in a centrifuge (no G suit) at 4G I was able to maintain consciousness at a pressure altitude of 30,000.  As soon as I was spun up to 6G things got ugly fast, and there is a video to prove it.
The operators would also have us spin at some G level and then explosively decompress the spinning chamber. LOC happens very fast in that case.
If one has the ability to adjust the Gs, it is possible to play with your oxygenation.  For example, performing the M1 maneuver, we would play with Gs and cause our vision to start to tunnel, to a soda straw tunnel, and then ease off on the Gs to get a more wider view of the world. The same thing is possible without the abdominal muscle maneuvers and working bladders in a G suit, just a lower Gs.

Is the video that proves it available on YouTube?

No, it's not on youtube.  It's kinda embarrassing to roll ones eyes up, just before one's neck gets floppy.  Hmm, I could put it on a limited access youtube, but I would first have to get it converted from beta.


text:
how can i help the air force at this age if I'm only 13 years old?
I'm young, 13 to be exact. I'm not old enough to join and I really want to help the brave men and women in the skies.How can I help the air force at this age?

You may add in which country you live and your nationality. Requirements to serve military one way or another differ from country and in each country depending on your nationality.

In many countries there's a kind of military league at school, perhaps high school. Maybe that would be good!\

Help them do what?  Is there anything in specific you would like to help them with?

As stated by Crossroads, AFROTC is an option. So is the Civil Air Patrol. Both will give you a taste of armed uniformed services life. Otherwise, look into volunteer opportunities locally to support the armed forces. Military.com may be a good start on researching charity organizations. Just be respectful in the forums, please.

In the US, you could look into Junior ROTC for the AF.
https://www.airforce.com/frequently-asked-questions/high-school/what-is-jrotc

Study.  Strive to become educated.  Hone your intelligence.  Learn science, engineering, and technology.  There are two reasons for this:

The Air Force is a technical service, dependent most upon enlistees who are strong in science and technology.  The air force prefers their enlistees to have degrees, and the more demanding jobs in the Air Force are filled by those with advanced degrees.  For example, test pilots tend to have advanced engineering degrees.

The Air Force is expensive to maintain and depends upon a fair amount of tax revenue.  If you do not enlist in the Air Force, but have developed strong engineering and science skills, you will have a higher than average salary that provides higher tax revenue to the government.  The government will have more tax revenue to buy the airplanes and other systems that make the Air Force work for those who did enlist.

Whether you enlist or not, strengthening your own mind and your own knowledge of science and technology will help the Air Force.

This, by a mile! First gain a broad perspective of life. Then focus. It's never the other way around. Keep the airforce in the back of your head, as a goal and an ideal. But to best help them, keep your eyes open.

In the UK, you can join the RAF Air Cadets at age 13 or older.
If you want a more civilian route at first, then there's also the Air Scouts

There's probably little that you (or anyone) can do to help the Air Force directly without going through the training required to join as service personnel.
There is, however, another route. You could contact the Air Force Aid Society, a charity that provides support to the men and women of the air force to help with their non-operational needs (education, healthcare, housing, etc.). They may have a branch local to you that would welcome your enthusiasm.
Alternatively, there are many organisations supporting service veterans who could use your help. Take a look at Charity Navigator to find organisations that do what you're interested in, and contact their local branches.


text:
Can I safely plug something that draws 6 amps into the cigar lighter (power port) in my Piper Archer lll?
Can I safely plug something that draws 6 amps into the cigar lighter (power port) in my Piper Archer lll ?

What's the breaker rating?

I wish I could find it. If it's in the POH, I don't see it.  Sent a note to my mechanic, hoping to get some info.

It depend of what you are trying to plug. If this is something not designed to withstand flight condition (pressure variation, acceleration,...), it is not safe to bring it in flight, plugged or not.

Yes, you can safely plug in a 6A load. Possibly a circuit breaker will pop out after some time. That depends on the rating of the circuit breaker serving the outlet and if there are any other loads on that circuit breaker. But it is completely safe to try it.

If you're going to "just try it", you may want to do so while taxiing for a while. Better that something unexpected go "pop" while still on the ground.

Cigar lighters usually have a 10amp breaker and wiring to match, but the sockets themselves are pretty poor quality and only designed to be active for 30 seconds or so. They can get pretty hot when used for any length of time, such as an air-bed inflator or tyre pump, even when the load is well below the breaker's limit.
If your use is at all critical, I'd be tempted to replace it with a bulkhead-mounted XT60 or similar connector.


text:
What was the normal flight path for a passenger aircraft from New York to Miami in 1955?
What was the normal flight path a passenger aircraft would have flown from New York to Miami in 1955? Like Pan American, flight 914.

Probably not a whole lot different than today. You likely won't get a very exact answer to a general question like that unless you can narrow down specifically what kind of detail you seek.
Obviously GPS waypoints wouldn't have been used, but any NDBs in use at the time probably would be. My advice would be to search for a vintage chart from that era and plot the most direct route using depicted navaids.


text:
What is the angle of Incidence of wing of Airbus A320?
What is the Angle of Incidence/ Mounting Angle of the wing of AIRBUS A320?

I vaguely recall it's around 5° at root to -0.5° at tip. The information is not particularly useful without the wing profile, which Airbus keeps to itself, and nothing is referenced to it; the stall angle of attack is referenced to the body axis.

@JanHudec General Aviation airplanes have about 6 deg. AOI. Isn't 5 deg. too much for commercial jets? Some websites suggest 1.5-2 deg. AOI but I don't know about the authenticity of this info.

I didn't say 5°, I said 5° at root to -0.5° at the tip. All swept wings are quite significantly twisted, otherwise they'd stall tip-first, which would be really, really nasty.

The A320 wing is a complex design with no clear mention of chord or angle of incidence, as it is not topical to the operation of the aircraft. The wing is swept back, has variable symmetry from root to tip , as some say twisted. This info is not shared by Airbus.


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Why don't bypass ducts in jet engines narrow down the flow to accelerate it?
Since decreasing the flow area increases flow velocity, why don't turbofan bypass ducts narrow down to produce more thrust?

How would accelerating the flow create more thrust?

It is really hard to say from cutaways  if the bypass duct narrow down.

@GdD A loss-less venturi maintains mass flow rate, then a larger differential between freestream speed and nozzle speed would increase thrust.

It increases the velocity of the air @JZYL, but it doesn't create more energy.

@GdD But it does create more thrust. See Asced's response below.

No, Asced says you get the same thrust either way @JZYL. 'there are two ways to produce the same thrust: strongly accelerate few air massflow or speed up a little bit a big amount of gas. Turbofans use the second strategy, mainly because it is more efficient.'

@GdD, for a given thrust which is an input to engine performance design (at some flight points), you have two “ways” indeed. The thrust formula is (in my opinion) a good explanation to show there is no need to accelerate the flow so much in nozzle. Nevertheless if you extend the nozzle, you still can accelerate the flow and so having more thrust (even if energy remains the same).

if you decrease the duct cross-section, you must apply work to accelerate the flow through the smaller duct. this means the fan driving the duct has to work harder (it must absorb more shaft horsepower) against that constriction in the flow and you derive no benefit- unless you mount the fan on a more powerful engine, in which case you are burning more fuel to drive the fan-and-duct system. So, there is no free lunch available here.

Flow acceleration/deceleration is isoenergetic/isentropic. The increase in dynamic pressure (acceleration) is balanced out by the decrease in static pressure as total pressure remains constant because no work is done on the fluid. Thrust should remain the same because you are trading momentum thrust (m_dot * v) for pressure thrust (p_e - p_ambient)*A_nozzle, however pressure thrust is not as efficient as momentum thrust, so a good nozzle will have a 0 pressure difference at design operating conditions.

Mostly because there is no need to strongly accelerate the flow. In fact, thrust engine is given with the formula below:
$$ F_N \approx \dot{m} \cdot \left(V_{out} - V_{in} \right) $$
Which means, there are two ways to produce the same thrust: strongly accelerate few air massflow or speed up a little bit a big amount of gas. Turbofans use the second strategy, mainly because it is more efficient.
Indeed, propulsion systems efficiency can be calculated as below, which highlights gains of creating thrust with small speed up and so huge massflow (bigger fan diameter, which is last decades tendency for civil aircrafts):
$$ \eta_{p} = \frac{F_N\cdot V_{in}}{\Delta \dot{E}_k} \approx \frac{\dot{m} \cdot \left(V_{out} - V_{in} \right)\cdot V_{in}}{\tfrac{1}{2}\dot{m}\left(V_{out}^2-V_{in}^2  \right)} = \frac{2}{1+\tfrac{V_{out}}{V_{in}}} $$

Note that this formula doesn't give you the overall engine efficiency resulting to thrust specific fuel (TSFC) consumption ($\eta_{thp}$ = $F_N$$V_{in}$ / $P_{fuel}$$\dot{m}_{fuel}$).

For a given engine, TSFC decrease if you add an extra duct in order to accelerate the flow, since more thrust is produced with same fuel. But, for a given thrust (fixed by aircraft flight mission) civil engine are globally more efficient with small acceleration.
Furthermore, the relatively small excess of thrust may not worth it due to extra weight needed to extend duct nozzle.

Turbofans use the second strategy, mainly because it is more efficient. - indeed, that's the point of building a high-bypass turbofan instead of a no-bypass turbojet.

Yes, completely agree

Yes, but “there are two ways to produce the same thrust” is perhaps a bit of a misleading way to phrase it. In fact, you always need to both have mass flow and accelerate it, the tradeoff decision is in how much of each you do.

@leftaroundabout: I thought it was clear with the next sentence. But I got your point, I could edit to avoid confusion

"For a given engine, TSFC decrease if you add an extra duct in order to accelerate the flow, since more thrust is produced with same fuel.",  If this is the case then you would make the aircraft more efficient. With more thrust, you could reduce the fuel flow back to the required level and end up with a better engine.

@ROIMaison. Yes you made the overall engine more efficient in the definition of TSFC (i.e. a little bit more thrust with same fuel). If the entire aircraft becomes more efficient ? I don't know because the extended nozzle involve extra weight in engines. Furthermore TSFC is not the only parameter to thing in an engine design, there is for instance, thermal / mechanical / noise / maintenance costs / operability / global weight / .. / considerations.

They do. It's just not by very much. Typical high bypass engine like a GE90 might narrow down the exit area by a few percent. Small enough that you probably don't even notice it when looking at it. But the air can feel it.

The simple answer is that the pressure ratio across fan is generally small, in order to maximize the propulsive efficiency, as outlined in @Asced's answer.
If you have a flow that has high total pressure (compared to inlet total pressure), then you have the real-estate to accelerate it further isentropically until the nozzle pressure reaches the ambient pressure, to maximize thrust.
However, if your total pressure behind the fan is not much higher than that of the inlet, then the available flow to accelerate via a convergent duct will be small. Depending on the pressure ratio, you can add a small ratio convergent duct, but you will need to trade off against viscous losses and added weight.

The physics are correct but the engines don't work this way, unfortunately. Most fans operate at axial Mach numbers around 0.5 (slowed down by the intake) -- if you were exhausting the bypass jet at M=0.5, you'd actually slow down the aircraft. The bypass nozzle does accelerate the flow, all the way to the speed of sound when the fan delivers enough pressure (which it does most of the time). I've put some pictures in my answer to show what that looks like.

Lots of turbofans have convergent nozzle fan ducts although the convergence is mild.  This is the Cf-34, which has a small but noticeable convergence.  On a lot of engines it's barely discernible.  It's because the designers are only after a small increase in velocity of the fan discharge.

The convergence is a consequence of the increase in diameter. The cross section area shouldn't change as much (but it does narrow down a bit; that much is correct).

Just a bit of taper in the 1st 3rd just after the compressor inlet.  Interesting thing about the CF-34. The annular compressor inlet in the earlier -3s was intended to be FOD resistant (a vestige of the A-10) and sees only a small ram effect and that plus the load from the hydraulic pump means these engines don't windmill very well and could core lock after a flameout if speed wasn't maintained. Later -#s raised the rear of the inlet lip to create a stronger ram effect into the compressor and partially unload the hydraulic pump so these engines will windmill right down to minimum flying speed.

Ok I get your point about the diameter.  There may be no effective choke at all.  If I was still working I could probably find out what the net cross sectional change is directly from GE.

You are of course correct in that higher exhaust velocity (for the same massflow) would produce more thrust. You may have overlooked that current turbofan engines do have a convergent bypass duct/nozzle. The axial Mach number at the fan is typically somewhere around Mach 0.5 to 0.6, and at the exhaust it needs to be much higher than that (because if you're flying at Mach 0.85, the flow needs to be faster than that, otherwise you have no jet...). However: The flow in the bypass duct will not accelerate indefinitely, since the power to accelerate it needs to come from somewhere...
Assuming subsonic flow (i.e. the flow stays below the speed of sound everywhere)
There is a certain ambient static pressure around the bypass exhaust nozzle*. That unfortunately already (mostly) determines the static pressure at the exit. The fan increases the total pressure of the flow (that's the pressure you would measure at the stagnation point of a probe), and the combination of total pressure in the bypass flow and the static pressure outside determines how far the bypass flow can accelerate. If you took a well-designed bypass duct and made the nozzle any smaller, you'd just get less massflow at the same exit velocity. In other words: The pressure ratio which the fan produces (minus intake and duct losses...) determines how much faster than the external flow the bypass flow can become when it exits the nozzle, and the amount of air you want to move through the duct determines the nozzle area. Now you could of course keep your nozzle area constant and make the fan bigger instead, but that would just mean you get slower flow through the fan, and a bigger, heavier fan. Which means more work for the intake to slow the flow down, a bigger engine, and not enough fan efficiency gained to make it worthwhile.
In transonic, compressible flow (i.e. what real passenger aircraft do these days)
Except for a few conditions (engines idle, taxi, partially during take-off -- for some particularly high-bypass engines maybe also at the start of climb), the pressure ratio of the fan is usually large enough to accelerate the flow to supersonic. Now as you may be aware, the cross-section of an accelerating streamtube contracts only until it reaches the speed of sound and expands afterwards. The theoretical ideal nozzle for this sort of thing is a convergent/divergent deNaval nozzle, and the massflow through such a nozzle is determined by the temperature, density and cross-section area at the narrowest point. This means that for a given desired massflow and a given operating condition (incoming flow, fan pressure ratio), the nozzle area is fixed.
In real life, most turbofans use a simple convergent nozzle to accelerate the flow from fan Mach number (M=0.5 to 0.6) to the speed of sound (M=1), or a convergent nozzle with a "flare" at the end. Here's a picture from this (unfortunately paywalled) paper of what the flow looks like:

You can nicely see how the bypass duct and nozzle are contracting and the flow is accelerating towards the exit, all the way to Mach 1, and how it continues to expand further and reach about Mach 1.3. (That's not a finished design, and the shock waves are stronger than they need to be. The paper goes on to explain how to reduce them but that's another topic).
The reasons why there's no convergent/divergent nozzle is that in slow flight (e.g. at take-off), the incoming air has fairly little dynamic pressure and the fan can't push the total pressure to supercritical by itself. Here's a picture (with a slightly different exhaust) at Mach 0.17:

A convergent/divergent nozzle here would incur bigger losses than a divergent only nozzle would in cruise, and take-off and early climb are critical design points for the engine. This, and not wanting to build a longer nozzle with a bigger exit radius are the reasons why turbofan engines (on civil airliners) don't usually have convergent/divergent nozzles.
Anyway: The point is that bypass ducts on modern turbofan engines do contract, and do accelerate the bypass flow, all the way to the speed of sound as long as the fan delivers enough pressure increase to allow that.
small aside on bypass nozzle design
One reason why underexpanding nozzles are not as bad as overexpanding ones is that the flow has more pressure than needed to get to Mach 1, and since supersonic flow expands, that's what it does as soon as it clears the nozzle (see the first picture). As a result, the streamlines in the back of the engine nacelle expand, and the jet takes up more space than if it was just nicely flowing out straight. That creates an "obstacle" for the outside flow which further increases static pressure on the exhaust system and also on the rear of the nacelle. This additional pressure pushes the aircraft forward. All in all, there's very little thrust lost, as long as the bypass jet is only a little faster than the speed of sound (as in this case). If it had enough pressure to get to Mach 2.0 or above, that would become a very different matter.
(*) to be more precise: The ambient static pressure around the nozzle is usually a bit above the farfield pressure. So the bypass jet does not accelerate all the way immediately but stays is a little slower on the rear of the engine and only quite gets to its final velocity a little behind the engine. You can see that nicely in the second picture, for the core jet which doesn't get to it's final Mach number until it passes the tip of the cone.

Exit pressure is not limited to the same as ambient - it is dictated by the flow through the engine. A well designed nozzle will have an exit pressure equal to ambient because pressure thrust is less efficient than momentum thrust. However, an aircraft operates through a variety of throttle settings and altitude, so it is expected that there will be pressure thrust and pressure drag at certain points of the flight envelope.

Also high bypass turbofans never have nozzles that are designed to go critical (M = 1). A convergent nozzle would generate shocks in the exhaust which would waste so much energy as well as destroying the hearing of anyone near the airport. Narcelles and inlets are designed to slow the incoming airflow so it always remains subsonic. even Concorde and SR-71 had mechanisms to ensure engine flow was always subsonic. Unless you have a fuel that combusts instantly, engine cores,and therefore bypasses must remain subsonic.

The actual pressure at the nozzle exit in flight is the same as around the nozzle exit. There's no way to avoid that. You might design your nozzle for some particular exit pressure, but get a different one in flight, but that just means that the actual velocity and massflow (and pressure produced by the fan) will adapt to the new counterpressure they're seeing. As I mentioned in the answer, underexpanded flow will expand outside of the nozzle and thus affect the exit pressure somewhat. Virtually all turbofans on airliners in service today are underexpanding in cruise.

The fan of course operates in subsonic axial flow. In the bypass duct itself, you'd try to keep the flow as slow as possible to minimize friction losses, but the nozzle contracts to accelerate the flow (because if you don't convert the pressure increase from the fan to jet velocity, why even have a fan?). Most turbofan engines (A320, not SR71!) don't have a divergent section, and the flow expands outside of the nozzle. Search for "underexpanded nozzle" to see pictures, but keep in mind that the bypass jet doesn't expand to Mach 3 but just ~1.3

I've looked for open-source pictures but the best thing I find is figure 3a on this paper (page 3):  https://mafiadoc.com/commercial-turbofan-engine-exhaust-nozzle-flow-_5b6bd551097c474c058b45e6.html    --  The "separate flow" configuration is what most recently-designed civil turbofans are using, and the figure shows some shock patterns on the oustide. However, those are pretty weak oblique shocks, and they're confined to the jet (of course). It shows the bypass jet expanding a little, too.

you are right about the exit pressure being ambient for subsonic nozzles, I completely forgot the Kutta condition. Supersonic flow is very different, and will have a component of pressure thrust if underexpanded. The expansion outside of the nozzle cannot affect the pressure at the nozzle exit - there is no feedback upstream in supersonic flow.

correct. The pressure at nozzle exit is the critical pressure which goes with the bypass flow total pressure, at least with "pure" convergent nozzles.

Regarding the lack of inefficiency due to underexpansion: the combination of the jets and the rear engine body looks very much like a toroidal aerospike to me. As long as the streamlines don't point outwards, I wouldn't expect much efficiency loss at all.

Thrust is not the whole story. Pure jets produce a very high velocity exhaust which may be suitable for a high speed aircraft such as a fighter, but airliners need to produce the required thrust at reasonable efficiency. Thrust is linearly proportional to momentum change of the air stream (linearly proportional to velocity change), but energy change is proportional to the square of the velocity change. What this means is that it is more energy efficient to make a small velocity change to a large mass of air than to make a large velocity change to a smaller mass of air. Processing large amounts of air can bring efficiency issues also, so a final design is an exercise in optimization trading off between numerous factors driven by the specific application (airliner, fighter, etc.).


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Where can I find electrical pinout of C152 microphone jack and DC bias?
Would anyone have the pinout and/or specs of the C152 passenger/pilot microphone jack? I'd also like to know what DC bias voltage, if any, the plane provides to the microphone.
Here's what I have so far, which may or may not be accurate:
              Headphone side     Plane Side
Sleeve             GND              GND
Ring         MIC OUT/BIAS IN       ???(?V)
Tip              UNUSED?            PTT?

I've dug through some Bendix/King nav/com radio manuals, but they don't tell you about what to expect on the plane side of things.
Thanks!

There are some great resources on webpages mostly aimed at home builders and experimental aviation enthusiasts as well as some electrical engineering forums.
Most of these aviation radios are set-up for use with a carbon microphone. They are fairly simple pieces of kit. If my memory serves me correctly the acceptable forward bias voltage range is set by the RTCA (https://www.rtca.org/) however I can't find a solid reference for this.
As far as I know, so called "GA plugs" (The 5.25mm/6.35mm style plugs) are wired as follows:
Sleeve             GND                           GND
Ring         MIC OUT/BIAS IN                    8-16v (Short circuit current ~150ma)
Tip       PTT  (Switch shorts to GND)            PTT

The reasoning behind the PTT switch still being assigned is for handheld backup microphones.
Many newer headsets don't use carbon microphones and use the electret capsule style microphones instead and therefore require some complex electrical engineering wizardry to drive.


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How can we physically define reynold number variation in terms of characteristic length?
How can you define it physically that
greater the characteristic length in
$Re=\frac{\rho v L_c}{\mu}$
greater the Reynolds number.
Like on wing surface $Re$ is higher than the horizontal tail surface (MiG 29).

So are you asking about the reynolds numbers, the way vortex generators work, the way dorsal fins and leading edge extensions work (similar to vortex generator, but not exactly the same) or the way reynolds number affects the later (not really)?

There are three questions in three questions in the post, sorry for the inconvenience i didn't write it up correctly. Vortex generators and Dorsal Fins how do they helps in delaying stall

No need to apologise, just edit the question to clarify what you are actually asking. I suggest you actually leave mention of the Reynolds number in, just explain why it makes the observed behaviour go against your intuition, so the answer can explain any bits you are missing.

If you have two independent questions, please make them as two separate posts (and I am pretty sure we have a detailed discussion of how vortex generators work already anyway).

Not sure what you're asking. Which part of "physically" are you confused about?

I am just asking why does reynold number changes from variation in characteristic length, Its physical explanation.

This question is not bad, as Reynolds number is determined by velocity, working fluid, and chord length.
Some designers try to make horizontal stabilizers miniature versions of their wings, resulting in safety issues.  Some even made the Hstab higher aspect, such as ones linking two fuselage "booms" together, abandoning the robust wedge shaped tails commonly seen on birds and biplanes of 100 years ago.
The result is a tail that stalls before its wing, not very good if you are diving towards earth.
Changing Reynolds number (of a smaller airfoil) can be counter-acted with change in shape to improve safety.  Note both the Virgin Galactic  "Knight" and the "Spaceship" now both sport rather Spitfire-like separate boom stabilizers.
As far as quantifying and understanding the effect of chord on airfoil properties at various Reynolds numbers, this would be an improvement in predicting the effects of changing Reynolds numbers (now seen on polar graphs).  Sort of like the Reynolds number as the molecule, and Velocity, Kinematic Viscosity, and Chord as its atoms.
Of these "atoms", Kinematic Viscosity is constant, chord and velocity are variable.  Effects of chord and Velocity changes are worth study in more detail.

The freestream Reynolds number ($Re_c$) of the wing and hstab would be different due to different chord lengths:
$$Re_c=\frac{\rho_\infty V_\infty c}{\mu}$$
where $\rho_\infty$ is the freestream air density, $V_\infty$ is the freestream airspeed and $c$ is the reference chord (of the wing or hstab). I'll discuss the physical significance of reference chord below.
1. Viscous drag
The physical significance on drag is best understood from the flat plate Blasius solution in laminar flow. The overall friction drag coefficient ($C_f$) decreases as a square root of freestream Reynolds number:
$$C_f=\frac{1.328}{\sqrt{Re_c}}$$
So the longer the chord (i.e. higher Re), the less significant the boundary layer viscous effects are to the overall drag. This is further illustrated in the graph below.

Image ref: https://www.sciencedirect.com/topics/engineering/friction-drag-coefficient
Take note, however, that the dimensionalized (total) friction drag still grows since $C_f$ here is non-dimensionalized with respect to the reference chord, $c$.
2. Turbulence transition
Increasing the chord doesn't increase the momentum thickness in absolute terms. However, the relative thickness with respect to the chord is different for different chord lengths:
$$\theta(s) = \frac{0.664\sqrt{\mu}}{\sqrt{\rho_\infty V_\infty}}s^{1/2}$$
where $s$ is the chordwise position from the leading edge.
Since natural transition is a function of the momentum thickness Reynolds number (Ref. Drela, Flight Vehicle Aerodynamics), which increases with respect to the momentum thickness, transition to turbulence will occur at similar chordwise location, $s$, in absolute terms, but earlier in relative terms ($s/c$) for the longer chord.
This has the effect of increasing the stall angle of attack for the larger chord, manifested in the effective increase in the freestream Reynolds number.


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What is the material of this tow bar?
What are the materials used in a tow bar of the pictured design? I am interested in the materials use for the tube (which aluminium alloy?) and the materials of the lunette mount flat and inside structure.

Have you tried asking the manufacturer?

I'm looking forward to the answer. My gut instinct is that material choice is not particularly sensitive because the application requires stiffness as the priority.  Typically, when stiffness is the goal, strength comes naturally. But typical != always.

Barium, of course.

Aircraft towbars are almost exclusively aircraft-grade aluminum or
  steel in construction  to meet or exceed the working load. Not only
  must the towbar be able to safely pull/push the vehicle, it must also
  be able to withstand the towed vehicle's inertia.

Source
Although there are several components that make up the tow bar they all must be strong enough to withstand the forces exerted on them. That's what the label on your picture is for, it dictates what aircraft you can use the tow bar on. Larger aircraft = Larger more robust tow bar.


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Why is Oiling Pneumatic Air Tools allowed in airplanes?
If I remember correctly oil in oxygen lines should be avoided due to the possibility of hazardous combustion. However I've seen recently that engineers use oil to lubricate pneumatic air tools. Can someone explain why this is allowed??

The oxygen lines are in the airplane.  The air tools use shop air.  From the shop compressor.  They are not related.  And nobody uses air tools around open oxygen lines anyway.

I edited the title to transform it into a question (this is a Q&A website) and to make it more specific. I had to assume your question was about airplanes. feel free to edit it again if it is not the case.

Perhaps better suited to the chemistry site, but air is only ~20% oxygen.  Diluting it with nitrogen makes combustion much more difficult.

This demonstrates the difference between oxygen and air.
Shop air is compressed air. It's about 20% oxygen. Compressed oxygen used in aviation is pure oxygen supplement, not compressed air, to save weight and volume of the gas you have to carry.
For breathing, the important thing is the partial pressure of oxygen, not the absolute oxygen content or the proportion of other gases.

shop air lines do not carry pure oxygen, they carry compressed air. any oil mist mixed into the compressed air will be far less flammable than oil mist mixed with oxygen.